[1] EXISTENCE FOR SOME VECTORIAL ELLIPTIC PROBLEMS WITH MEASURE DATA 33
F
RANCESCOL
EONETTI andP
IERV
INCENZOP
ETRICCA(*)
Existence for some vectorial elliptic problems
with measure data (**)
1 - Introduction
The study of elliptic boundary value problems with measure data div (a(x;u(x);Du(x)))m inV
u0 on@V
(
1:1
received great attention starting from[2]. In such a paper the scalar case
u:VRn !Rhas been considered and the model operator is the p-laplacian
ai(x;u;j) jjjp 2ji. Some years later [3], the anisotropic pi-laplacian has been dealt with:ai(x;u;j) jjijpi 2ji, that is, each component of the gradientDiumay have a different exponent pi; this seems useful when dealing with some re-inforced materials, [11]; see also [10], example 1.7.1, page 169. Let us point out that both [2] and [3] deal with the scalar case u:V!R; as far as the vectorial caseu:V!RNis concerned, some contributions appeared in [7] and [4], where the vectorialp-laplacian is considered. In this paper we deal with the anisotropic
(*) F. Leonetti: Dipartimento di Matematica Pura ed Applicata, UniversitaÁ di L'Aquila, 67100 L'Aquila, Italy; e-mail: [email protected]; P.V. Petricca: Via Sant'Amasio 18, 03039 Sora, Italy.
(**) Received November 17th2005 and in revised formMarch 2nd2006. AMS classifica-tion 35J60, 35D05, 35J55.
vector valued case: Pn i1Di(jDiuj pi 2D iu)m inV u0 on@V; 8 < : 1:2
where u:V Rn!RN with n3, N1, V is a bounded open set and
m(m1;. . .;mN) is a finite Radon measure onRnwith values intoRN. In this paper we prove existence of weak solutions for the Dirichlet problem(1.2): in Section 2 we give the precise result, whose proof appears in Section 3. We thank the referee for useful remarks and suggestions.
2 - Notations and statements
We study existence of weak solutions for the Dirichlet problem
Pn i1Di(jDiuj pi 2D iu)m inV u0 on@V; 8 < : 2:1
where u:V Rn!RN with n3, N1, V is a bounded open set and
m(m1;. . .;mN) is a finite Radon measure onRnwith values intoRN. We need some restrictions on the exponentspi2; we introduce the harmonic mean
p n1Xn i1
1
pi
! 1
and we assume that
p<nand p(p(n 1)1)n<pi<p(n pn 1) 8i2 f1;. . .;ng: 2:2
The right hand side of (2.2) can be written as
pi 1 np((np 1)1)
pipn p(n 1)<1 thus
1pi 1<np((pn 1)1)pi so we can take exponentsq1;. . .;qnsuch that
and we set qmin i qi; q 1 n Xn i1 1 qi ! 1 ; q11q n1:
In this paper we will prove the following
Theorem2.1. Under the previous assumptions there exists u2W01;q(V;RN) \Lq(V;RN)with D iu2Lqi(V;RN) 8i2 f1;. . .;ngsuch that V Xn i1 XN b1 jDiujpi 2DiubDiWbdLn XN b1 V Wbdmb 8W2C1 c (V;RN): 2:3
In this theoremwe may taken3,p1p22 andp32(2;1); this example shows that our theoremcannot be derived from[5], [13].
The previous theoremimproves on [9]: we remove the restriction on the support of the measuremand we allow a wider range forpi.
3 - Proof of the Theorem
LetfQkgk2Nbe a decreasing sequence of positive numbers converging to zero. We mollify our measuremand we obtain smooth functions
fkmrQk* m with sup k kfkkL1(V) jmj(R n)<1: 3:1
We setpminipiand we use direct methods in the calculus of variations in order to find uk2 V n u2W01;p(V;RN): Diu2Lpi(V;RN) 8i2 f1;. . .;ng o ; such that V Xn i1 XN b1 jDiukjpi 2DiubkDiWbdLn XN b1 V Wbfb kdLn 8W2 V: 3:2
on level sets [2]; we set M(t) 0 0tL t L L<t<L1 1 tL1 M( t) t<0 8 > > > > < > > > > :
and we use the following test functionWin (3.2): we fixa2 f1;. . .;Ngand we take
W(W1;. . .;WN) withWb0 forb6aandWa M(ua
k), whereuakis thea-th compo-nent ofuk(u1k;. . .;uNk). We define
BL;kx2V:Luak(x)<L1 ; and we insert the previousWinto (3.2): we get
BL;k Xn i1 Diukpi 2D iuak2dx V fa kM(uak)dx: 3:3
SincejM(t)j 1, we can use (3.1) so that the right hand side of (3.3) is bounded by jmj(Rn). Let us look at the left hand side: sincepi2, we have
Diua kpi 2Diukpi 2 thus we get BL;k Xn i1 Diua kpidxc1 8k;L2N: 3:4 Since qi<np((np 1)1)pi; we can select u2 0;n(p 1) p(n 1) (0;1)
such thatqiupifor everyi. Now, without loss of generality, we can assume that
qiupi for everyiand
Let us remark that 1<qi<pifor everyi2 f1;. . .;ngandq<p<n. Now we follow [3]: we set
l(1 uu)q
so thatl>1; we use Holder inequality:
V Diua kqidx V Diua kqi(1 juakj) lqi=pi(1 juakj)lqi=pidx V Diua kpi(1 juakj) ldx 2 4 3 5 qi=pi V (1 jua kj)lqi=(pi qi)dx 2 4 3 5 1 qi=pi X1 L0 BL;k Diua kpi(1 juakj) ldx 2 6 4 3 7 5 qi=pi V (1 jua kj)lqi=(pi qi)dx 2 4 3 5 1 qi=pi X1 L0 (1L) l BL;k Diua kpidx 2 6 4 3 7 5 qi=pi V (1 jua kj)q dx 2 4 3 5 1 qi=pi : 3:5
We keep in mind (3.4) andl>1: we get
V Diua kqidxc2 V (1 jua kj)q dx 2 4 3 5 1 qi=pi : 3:6
We use the anisotropic embedding theorem [12] and we get
ua kLq(V)c3 Yn i1 Diua k1L=qin(V)c4 Yn i1 V (1 jua kj)q dx 2 4 3 5 [(1=qi) (1=pi)]=n c4 V (1 jua kj)q dx 2 4 3 5 (1 u)=(up) : 3:7 We recall thatp<n, so (1 u)=(up)<1=q thus (3.7) gives ua kLq(V)c5 8k2N; 8a1;. . .;N: 3:8
We insert this bound into (3.6) and we get
Diua
kLqi(V) c6 8k2N; 8i1;. . .n; 8a1;. . .;N: 3:9
Because of weak compactness, there exists u2W01;q(V;RN)\Lq(V;RN), with
Diu2Lqi(V;RN) 8i2 f1;. . .;ng, si2L
qi
pi 1(V;RN) and h
i2Lqi(V;R) for every
i2 f1;. . .;ngsuch that, up to a subsequence,
uk*u in W01;q(V;RN); 3:10 uk !u a:e:inV; 3:11 Diuk*Diu inLqi(V;RN) 8i2 f1;. . .;ng; 3:12 jDiukjpi 2Diuk*si inL qi pi 1(V;RN) 8i2 f1;. . .;ng; 3:13 jDiuk Diuj*hi0 inLqi(V;R) 8i2 f1;. . .;ng: 3:14 We claimthat hi0 inV 8i2 f1;. . .;ng 3:15 thus Diuk!Diu inL1(V;RN) 8i2 f1;. . .;ng: 3:16
We prove claim(3.15) following [4]. We fix a smoothv:Rn!RN; later, vwill be chosen as a linear Taylor polynomial ofu. We take cut off functionsf:Rn!Rand
h:RN!Rsuch that
f2C1
c (V;R);
h2C1
c (RN;R);
with 0f1 and 0h1. We takec:RN!RNas follows
c(z)g(jzj)jzzj
where g2C1((0;1);R) with g and g0 non-negative and bounded; moreover, we require thatcjsupp (h) Id . Note that
@ca @zs (z)g0(jzj) zsza jzj2 g(jzj) jzj dsa zsza jzj2 ! thus XN a1 jaXN s1 @ca @zs(z)js0 8z;j2RN: 3:17
For everyj2 f1;. . .;ngit results that 0 V jDjuk Djujh(uk v)fdx V jDjuk Djvjh(uk v)fdx V jDjv Djujh(uk v)fdx V jDjuk Djvjpjh(uk v)fdx 0 @ 1 A 1=pj V h(uk v)fdx 0 @ 1 A 1 1=pj V jDjv Djujh(uk v)fdx: 3:18
Let~pmaxipi. Then, using the monotonicity ofz! jzjp 2zand the equalitycId on supp(h), we get 0Tkj : V jDjuk Djvjpjh(uk v)fdx Xn i1 V jDiuk Divjpih(uk v)fdx 8~p 1Xn i1 V hjDiukjpi 2Diuk jDivjpi 2Div;Diuk Divih(uk v)fdx 8~p 1Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v))ih(uk v)fdx 8~p 1Xn i1 V hjDivjpi 2Div;Diuk Divih(uk v)fdx 8~p 1 Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v))ifdx 8 < : Xn i1 V hjDivjpi 2Div;Diuk Divih(uk v)fdx Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v))i[1 h(uk v)]fdx 9 = ;: 3:19
We recall (3.2) and we get Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v))ifdx Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v)f)idx Xn i1 V hjDiukjpi 2Diuk;c(uk v)Difidx V hmrQk;c(uk v)fidx Xn i1 V hjDiukjpi 2Diuk;c(uk v)Difidx: 3:20 Moreover, (3.17) implies Xn i1 V hjDiukjpi 2Diuk;Di(c(uk v))i[1 h(uk v)]fdx Xn i1 V XN a1 jDiukjpi 2Diuak XN s1 @ca @zs(uk v)Divs[1 h(uk v)]fdx: 3:21
We insert (3.20) and (3.21) into (3.19): we get 0Tjk: V jDjuk Djvjpjh(uk v)fdx 8~p 1 V hmrQk;c(uk v)fidx 8 < : Xn i1 V hjDiukjpi 2Diuk;c(uk v)Difidx Xn i1 V XN a1 jDiukjpi 2Diuak XN s1 @ca @zs(uk v)Divs[1 h(uk v)]fdx Xn i1 V hjDivjpi 2Div;Diuk Divih(uk v)fdx 9 = ;: 3:22
continuous and bounded: for everyj1;. . .;nwe get 0Tj 1:limsup k!1 T j k 8~p 1 sup z2RNjc(z)j V fdjmj Xn i1 V hsi;c(u v)Difidx 8 < : Xn i1 V XN a1 sa i XN s1 @ca @zs(u v)Divs[1 h(u v)]fdx Xn i1 V hjDivjpi 2Div;Diu Divih(u v)fdx 9 = ;; 3:23 0 lim k!1 V jDjuk Djujh(uk v)fdx V hjh(u v)fdx; 3:24 lim k!1 V jDjv Djujh(uk v)fdx V jDjv Djujh(u v)fdx; 3:25 lim k!1 V h(uk v)fdx V h(u v)fdx: 3:26
Take the limsup in (3.18). We get, for everyj1;. . .;n, 0 V hjh(u v)fdx Tj 1 1=pj V h(u v)fdx 0 @ 1 A 1 1=pj V jDjv Djujh(u v)fdx 3:27 whereTj
1 has been estimated in (3.23). For a. e.a2Vand everyj1;. . .;n, we have (see [6] and [1])
lim R!0 B(a;R) ju(x) [u(a)Du(a)(x a)]j R dx0; 3:28 lim R!0 B(a;R) jDu(x) Du(a)jdx0; 3:29 lim R!0 B(a;R) jsj(x) sj(a)jdx0; 3:30
lim R!0 B(a;R) jhj(x) hj(a)jdx0; 3:31 limsup R!0 jmj(B(a;R)) Ln(B(a;R))<1 3:32
whereLn is the Lebesgue measure inRnand
A f(x)dxLn1(A) A f(x)dx:
We choosevto be the linear Taylor polynomial ofuina:
v(x)u(a)Du(a)(x a): Since we are still free to selectf,handc, let
fR(x)R1nf~x aR ; ~f2C1 c (B(0;1);R); 0~f1; Rn ~ f(x)dx1 hR(y)~h Ry ; ~h2C1c (B(0;1);R); 0h~1; ~hjB(0;1 2)1; cR(y)R~c Ry ;
withc~ as in the above discussion. Then (3.27) can be written as follows: 0 B(a;R) hjhR(u v)fRdx Tj1;R1=pj B(a;R) hR(u v)fRdx 0 B @ 1 C A 1 1=pj B(a;R) jDjv DjujhR(u v)fRdx 3:33 where 0Tj1;R:limsup k!1 B(a;R) jDjuk DjvjpjhR(u v)fRdx 8~p 1 ( sup z2RNjcR(z)j B(a;R) fRdjmj Xn i1 B(a;R) hsi;cR(u v)DifRidx Xn i1 B(a;R) XN a1 sa i XN s1 @ca R @zs (u v)Divs[1 hR(u v)]fRdx Xn i1 B(a;R) hjDivjpi 2Div;Diu DivihR(u v)fRdx ) : 3:34
Recalling (3.28), ... , (3.32) and the definitions ofhR,fR, we get lim R!0 B(a;R) hjhR(u v)fRdxhj(a); 3:35 lim R!0T j 1;R0; 3:36 0 B(a;R) hR(u v)fRdx1; 3:37 lim R!0 B(a;R) jDjv DjujhR(u v)fRdx0: 3:38
As far as (3.36) is concerned, we give some details in the Appendix. Let us take
R!0in (3.33): we get, for everyj1;. . .;n,
hj(a)0 for a.e:a2V 3:39
thus our claim(3.15) is proved. Now we have (3.16), thus, up to a subsequence,
Duk!Dua:e:in V: 3:40
The previous pointwise convergence and the estimate
jDiukjpi 2D iuk Lpiqi1(V)Diuk pi 1 Lqi(V)c7; 8k2N; 8i2 f1;. . .;ng allow us to get jDiukjpi 2Diuk*jDiujpi 2Diu in L qi pi 1(V;RN) 8i2 f1;. . .;ng: 3:41
Thus we can pass to the limit, ask! 1, in (3.2) and it turns out thatusolves (2.3).
This ends the proof. p
4 - Appendix
For the convenience of the reader, we give some details about the proof of (3.36). We look at inequality (3.34): it suffices to prove that all the four integrals go to zero whenR!0. We keep in mind that
jcR(z)j c8R 4:1
0fR(x)R1n 4:2
then 0 sup z2RNjcR(z)j B(a;R) fRdjmj c8Rjmj(BR(an;R))
and the right hand side goes to zero because of (3.32). Now we deal with the second integral in the right hand side of (3.34):
B(a;R) hsi(x);cR(u(x) v(x))DifR(x)idx B(a;R) hsi(a);cR(u(x) v(x))DifR(x)idx B(a;R) hsi(x) si(a);cR(u(x) v(x))DifR(x)idx IRIIR: Note that DifR(x)R1n Di~f x a R h i1 R: We recall that ~ c(z)zforz2B 0;12 supp (~h) then cR(u v)u v on ju vR j12
thus it is convenient to let
AR x2B(a;R):ju(x)Rv(x)j>12
:
We definevnto be thendimensional Lebesgue measure of the unit ball inRn; thenLn(B(a;R))v nRnand 0LnR(AnR)2vn B(a;R) ju(x) v(x)j R dx:
We use (3.28) and we get lim R!0 Ln(AR) Rn 0: 4:3
Now we are able to deal withIRas follows:
0IR R1n AR hsi(a);cR(u(x) v(x)) Di~f x aR h i1 Ridx 1 Rn B(a;R)nAR hsi(a);cR(u(x) v(x)) Dif~ x aR h i1 Ridx 4:4 LnR(AnR)jsi(a)jc8Dif~L1(Rn)vnjsi(a)jDi~fL1(Rn) B(a;R) ju(x) v(x)j R dx:
Using (4.3) and (3.28) we get
lim
R!0IR0: 4:5
As far asIIRis concerned, we argue as follows: 0IIRc8vnDif~L1(Rn)
B(a;R)
jsi(x) si(a)jdx
thus we can use (3.30) and we get lim
R!0IIR0: 4:6
Limits (4.5) and (4.6) guarantee that the second integral in (3.34) goes to zero when
R!0. In a similar manner we can show that the third integral goes to zero. The fourth integral is easier to be dealt with. Thus we have established (3.36).
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Abstract
We study the Dirichlet problem
Pn i1Di(jDiuj pi 2D iu)m inV u0 on@V; 8 < :
whereu:VRn!RNwithn3,N1,Vis a bounded open set andm(m1;. . .;mN)is a
finite Radon measure onRnwith values intoRN. Under some restrictions on the exponents
pi2, we prove existence of a weak solutionu.